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Chapter 9: Problem 59

A circular saw blade with radius \(0.120 \mathrm{~m}\) starts from rest and turns in a vertical plane with a constant angular acceleration of \(2.00 \mathrm{rev} / \mathrm{s}^{2} .\) After the blade has turned through \(155 \mathrm{rev},\) a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of \(0.820 \mathrm{~m}\) to the floor. How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?

Short Answer

Expert verified
The horizontal distance covered by the piece is obtained by substituting the known values into the range equation.

Step by step solution

01

Convert angular acceleration and displacement to linear

Use the relationships that link angular and linear quantities. For linear velocity \(v\), this is done by multiplying the angular velocity by the radius: \(v = \omega r\). The same rule applies for acceleration: \(a = \alpha r\). In this case, we convert the given angular acceleration (in rev/s\(^2\)) to rad/s\(^2\) by multiplying with \(2\pi\). This gives \(a = \alpha r = 2 \cdot 2\pi r\). Similarly convert the angular displacement of 155 revolutions to radians by multiplying by \(2\pi\). This gives a displacement, \(d = 155 \cdot 2\pi r\).
02

Find the final velocity

Use the equation of motion that relates displacement, initial velocity, acceleration and final velocity: \(v_f^2 = v_i^2 + 2ad\). Here initial velocity \(v_i\) is zero (starts from rest), acceleration \(a\) is the one found in step 1 and displacement \(d\) is also found in step 1. Solve for \(v_f\) to get the final (tangential) velocity of the blade and hence the initial speed of the piece.
03

Establish the time of flight

Use the kinematic equation for vertical motion: \(h = v_{iy}t + 0.5g t^2\), where \(v_{iy}\) is initial vertical velocity, 0 in our case since the piece departs horizontally; \(h\) is the height from which the piece fell, given as 0.820m, \(g\) is the acceleration due to gravity, and \(t\) is the time it takes for the piece to hit the ground. Solve for \(t\) to find the time of flight.
04

Calculate horizontal distance covered

The horizontal motion is uniform, so the horizontal distance or range \(R\) covered by the piece is given by: \( R = v_{ix} t\), where \(v_{ix}\) is the initial horizontal velocity, equal to the final tangential velocity of the blade found in Step 2 because the piece broke off tangentially, and \(t\) is the time of flight determined in Step 3. Solve for \(R\) to get the answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Motion
Angular motion describes how an object rotates around a fixed point or axis. It is characterized by quantities such as angular velocity and angular acceleration. These quantities define how quickly and how much the object speeds up during its rotation.
  • **Angular Velocity**: This is the rate at which an object rotates, usually measured in revolutions per second or radians per second. In the exercise, the angular velocity is initially zero because the blade starts from rest.
  • **Angular Acceleration**: This is the rate of change of angular velocity, described as revolutions per second squared (rev/s虏). Here, it鈥檚 given as 2.00 rev/s虏.
  • **Displacement**: Measures how much rotation has occurred, in revolutions or radians. For the exercise, displacement is 155 revolutions, which needs to be converted to radians for calculation.
Understanding these concepts allows us to relate angular quantities to linear motion through the radius of rotation, as seen when converting angular acceleration to linear acceleration.
Kinematics
Kinematics is the branch of mechanics that deals with motion without considering the forces that cause it. Here we explore how kinematics applies to both linear and angular motion.
  • **Equation of Motion**: Kinematics uses equations to relate velocity, acceleration, time, and displacement. The primary equation used here, \(v_f^2 = v_i^2 + 2ad\), allows us to find the final linear velocity from rest, which is crucial in solving for how far the piece travels horizontally.
  • **Time of Flight**: Since the piece of the blade drops freely under gravity, we use the kinematic equation for vertical motion \(h = v_{iy}t + 0.5g t^2\) to determine the time it takes to hit the ground. This is essential for calculating horizontal travel.
  • **Vertical Displacement**: The vertical distance the piece travels is a key factor, given as 0.820 meters, providing the necessary height variable for our kinematic equation.
By mastering these kinematic principles, we can efficiently navigate projectile motion problems like the one presented in this exercise.
Circular Motion
In circular motion, an object revolves around a central point. The path of the motion is a circle, and various physical laws and formulas apply, involving the radius as a crucial factor.
  • **Tangential Velocity**: Comprehending circular motion requires understanding tangential velocity (\( v = \omega r \)). This represents the linear velocity at any point along the edge of the circular path. For our problem, this determines the initial horizontal speed of the piece as it breaks off the blade.
  • **Centripetal Acceleration**: Although not directly addressed here, it often plays a role in circular motion, meeting the requirements for objects to maintain a circular path. It's calculated using \( a_c = v^2/r \).
  • **Relation to Linear Motion**: Circular motion concepts convert to linear motion through the radius of the circle. In this way, angular displacement, velocity, and acceleration convert to linear equivalents that are directly applicable in projectile motion once the object leaves its circular path.
Leading to projectile path analysis, these circular motion insights are valuable in comprehending how rotational dynamics transition to linear dynamics.

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Most popular questions from this chapter

The rotating blade of a blender turns with constant angular acceleration \(1.50 \mathrm{rad} / \mathrm{s}^{2}\). (a) How much time does it take to reach an angular velocity of \(36.0 \mathrm{rad} / \mathrm{s},\) starting from rest? (b) Through how many revolutions does the blade turn in this time interval?

An airplane propeller is \(2.08 \mathrm{~m}\) in length (from tip to tip) with mass \(117 \mathrm{~kg}\) and is rotating at 2400 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod. (a) What is its rotational kinetic energy? (b) Suppose that, due to weight constraints, you had to reduce the propeller's mass to \(75.0 \%\) of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?

You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius \(25.0 \mathrm{~cm} .\) Starting from rest at \(t=0,\) the flywheel rotates with constant angular acceleration \(3.00 \mathrm{rad} / \mathrm{s}^{2}\) about an axis perpendicular to the flywheel at its center. If the flywheel has a density (mass per unit volume) of \(8600 \mathrm{~kg} / \mathrm{m}^{3},\) what thickness must it have to store \(800 \mathrm{~J}\) of kinetic energy at \(t=8.00 \mathrm{~s} ?\)

A child is pushing a merry-go-round. The angle through which the merry-go- round has turned varies with time according to \(\theta(t)=\gamma t+\beta t^{3}, \quad\) where \(\quad \gamma=0.400 \mathrm{rad} / \mathrm{s} \quad\) and \(\quad \beta=0.0120 \mathrm{rad} / \mathrm{s}^{3}\) (a) Calculate the angular velocity of the merry-go-round as a function of time. (b) What is the initial value of the angular velocity? (c) Calculate the instantaneous value of the angular velocity \(\omega_{z}\) at \(t=5.00 \mathrm{~s}\) and the average angular velocity \(\omega_{\mathrm{av}-z}\) for the time interval \(t=0\) to \(t=5.00 \mathrm{~s}\) Show that \(\omega_{\mathrm{av}-z}\) is \(n o t\) equal to the average of the instantaneous angular velocities at \(t=0\) and \(t=5.00 \mathrm{~s},\) and explain.

The angular velocity of a flywheel obeys the equation \(\omega_{z}(t)=A+B t^{2},\) where \(t\) is in seconds and \(A\) and \(B\) are constants having numerical values 2.75 (for \(A\) ) and 1.50 (for \(B\) ). (a) What are the units of \(A\) and \(B\) if \(\omega_{z}\) is in \(\mathrm{rad} / \mathrm{s} ?\) (b) What is the angular acceleration of the wheel at (i) \(t=0\) and (ii) \(t=5.00 \mathrm{~s} ?\) (c) Through what angle does the flywheel turn during the first 2.00 s? (Hint: See Section \(2.6 .)\)
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